# Prove this? sinh(a+b)=sinhacoshb+sinhbcosha

Apr 8, 2018

Main Identities i'll be using here are,
color(red)(cosh x =(e^x + e^-x)/2 color(white)(rrr and color(white)(rrr color(magenta)(sinh x=(e^x-e^-x)/2

To Prove $\to$$\sinh \left(a + b\right) = \sinh a \cosh b + \sinh b \cosh a$

I'll begin with the right side,
color(white)(rrr

=>color(magenta)(sinha)color(red)(coshb)+color(magenta)(sinhb)color(red)(cosha

$\implies \left(\textcolor{m a \ge n t a}{\frac{{e}^{a} - {e}^{-} a}{2}} \times \textcolor{red}{\frac{{e}^{b} + {e}^{-} b}{2}}\right) + \left(\textcolor{m a \ge n t a}{\frac{{e}^{b} - {e}^{-} b}{2}} \times \textcolor{red}{\frac{{e}^{a} + {e}^{-} a}{2}}\right)$

color(white)(rrr

$\implies \frac{\left({e}^{a} - {e}^{-} a\right) \left({e}^{b} + {e}^{-} b\right)}{4} + \frac{\left({e}^{b} - {e}^{-} b\right) \left({e}^{a} + {e}^{-} a\right)}{4}$

color(white)(rrr

$\implies \frac{1}{4} \left[{e}^{a + b} + {e}^{a - b} - {e}^{b - a} - {e}^{- \left(a + b\right)} + {e}^{a + b} - {e}^{a - b} + {e}^{b - a} - {e}^{- \left(a + b\right)}\right]$

color(white)(rrr

$\implies \frac{1}{4} \left[{e}^{a + b} + \cancel{{e}^{a - b}} - \cancel{{e}^{b - a}} - {e}^{- \left(a + b\right)} + {e}^{a + b} \cancel{- {e}^{a - b}} \cancel{+ {e}^{b - a}} - {e}^{- \left(a + b\right)}\right]$

color(white)(rrr

$\implies \frac{1}{4} \left[2 {e}^{A + B} - 2 {e}^{- \left(A + B\right)}\right]$

color(white)(rrr

=>1/cancel4^2[cancel2(e^(A+B)-e^(-(A+B))]

color(white)(rrr

$\implies \frac{{e}^{A + B} - {e}^{- \left(A + B\right)}}{2}$

color(white)(rrr

$\implies \sinh \left(A + B\right)$

Hence Proved :)